2次の偏微分

■問題

次の関数の第2次偏導関数を求めよ．

$3$\displaystyle{z=\log $ x- y $ }$

■答

$3$\displaystyle{\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{x}^{2} \hspace{2}}=- \frac{\hspace{2}1\hspace{2}}{\hspace{2} { $ x- y $ }^{2} \hspace{2}}}$ ， $3$\displaystyle{\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{y}^{2} \hspace{2}}=- \frac{\hspace{2}1\hspace{2}}{\hspace{2} { $ x- y $ }^{2} \hspace{2}}}$ ， $3$\displaystyle{\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ \frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}x \hspace{2}} $ =\frac{\hspace{2}1\hspace{2}}{\hspace{2} { $ x- y $ }^{2} \hspace{2}}}$

■ヒント

偏導関数の定義を用いて2次の偏微分を行う．

■解説

偏導関数の定義より， $3$\displaystyle{\text{\hspace{1}}\text{\hspace{1}}y\text{\hspace{1}}\text{\hspace{1}}}$ を定数とみなして $3$\displaystyle{\text{\hspace{1}}\text{\hspace{1}}x\text{\hspace{1}\hspace{1}}}$ で微分する．

$3$\displaystyle{\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}x \hspace{2}}}$ $3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2} x- y \hspace{2}}\text{\hspace{1}}\cdot \text{\hspace{1}}\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}} $ x- y $ }$

$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2} x- y \hspace{2}}\text{\hspace{1}}\cdot \text{\hspace{1}}1}$

$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2} x- y \hspace{2}}}$

$3$\displaystyle{={ $ x- y $ }^{ - 1 }}$

これを更に $3$\displaystyle{\text{\hspace{1} }\text{\hspace{1} }x\text{\hspace{1} }\text{\hspace{1} }}$ で偏微分すると，

$3$\displaystyle{\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{x}^{2} \hspace{2}}}$ $3$\displaystyle{=\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}} $ \frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}x \hspace{2}} $ }$

$3$\displaystyle{=\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}} \{ { $ x- y $ }^{ - 1 } \} }$

$3$\displaystyle{=- 1\text{\hspace{1}}\cdot \text{\hspace{1}}{ $ x- y $ }^{ - 2 }\text{\hspace{1}}\cdot \text{\hspace{1}}\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}} $ x- y $ }$

$3$\displaystyle{=- { $ x- y $ }^{ - 2 }\text{\hspace{1}}\cdot \text{\hspace{1}}1}$

$3$\displaystyle{=- { $ x- y $ }^{ - 2 }}$

$3$\displaystyle{=- \frac{\hspace{2}1\hspace{2}}{\hspace{2} { $ x- y $ }^{2} \hspace{2}}}$

同様の手順で偏導関数の定義より， $3$\displaystyle{\text{\hspace{1}}\text{\hspace{1}}x\text{\hspace{1}}\text{\hspace{1}}}$ を定数とみなして $3$\displaystyle{\text{\hspace{1}}\text{\hspace{1}}y\text{\hspace{1}}\text{\hspace{1}}}$ で微分する．

$3$\displaystyle{\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}y \hspace{2}}}$ $3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2} x- y \hspace{2}}\text{\hspace{1}}\cdot \text{\hspace{1}}\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ x- y $ }$

$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2} x- y \hspace{2}}\text{\hspace{1}}\cdot \text{\hspace{1}} $ - 1 $ }$

$3$\displaystyle{=- \frac{\hspace{2}1\hspace{2}}{\hspace{2} x- y \hspace{2}}}$

$3$\displaystyle{=- { $ x- y $ }^{ - 1 }}$

$3$\displaystyle{\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{y}^{2} \hspace{2}}}$ $3$\displaystyle{=\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ \frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ }$

$3$\displaystyle{=\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}} \{ - { $ x- y $ }^{ - 1 } \} }$

$3$\displaystyle{=- 1\text{\hspace{1}}\cdot \text{\hspace{1}}- { $ x- y $ }^{ - 2 }\text{\hspace{1}}\cdot \text{\hspace{1}}\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ x- y $ }$

$3$\displaystyle{={ $ x- y $ }^{ - 2 }\text{\hspace{1}}\cdot \text{\hspace{1}} $ - 1 $ }$

$3$\displaystyle{=- { $ x- y $ }^{ - 2 }}$

$3$\displaystyle{=- \frac{\hspace{2}1\hspace{2}}{\hspace{2} { $ x- y $ }^{2} \hspace{2}}}$

また，２次偏導関数は偏微分する順序には無関係であるから， $3$\displaystyle{\text{\hspace{1} }\text{\hspace{1} }z\text{\hspace{1} }\text{\hspace{1} }}$ を $3$\displaystyle{\text{\hspace{1} }\text{\hspace{1} }x\text{\hspace{1} }\text{\hspace{1} }}$ で偏微分した後 $3$\displaystyle{\text{\hspace{1} }\text{\hspace{1} }y\text{\hspace{1} }\text{\hspace{1} }}$ で偏微分すると，

$3$\displaystyle{\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}x \hspace{2}}\text{\hspace{1}}}$ $3$\displaystyle{={ $ x- y $ }^{ - 1 }}$

$3$\displaystyle{\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ \frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}x \hspace{2}} $ }$ $3$\displaystyle{=\frac{\hspace{2} {{\partial}}^{2} \hspace{2}}{\hspace{2} {\partial}y{\partial}x \hspace{2}}}$

$3$\displaystyle{=\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}} \{ { $ x- y $ }^{ - 1 } \} }$

$3$\displaystyle{=- { $ x- y $ }^{ - 2 }\text{\hspace{1}}\cdot \text{\hspace{1}} $ - 1 $ }$

$3$\displaystyle{={ $ x- y $ }^{ - 2 }}$

$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2} { $ x- y $ }^{2} \hspace{2}}}$ 　　･･････(1)

$3$\displaystyle{\text{\hspace{1} }\text{\hspace{1} }z\text{\hspace{1} }\text{\hspace{1} }}$ を $3$\displaystyle{\text{\hspace{1} }\text{\hspace{1} }y\text{\hspace{1} }\text{\hspace{1} }}$ で偏微分した後 $3$\displaystyle{\text{\hspace{1} }\text{\hspace{1} }x\text{\hspace{1} }\text{\hspace{1} }}$ で偏微分すると，

$3$\displaystyle{ \frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}y \hspace{2}}=- { $ x- y $ }^{ - 1 } }$

$3$\displaystyle{\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}} $ \frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ }$ $3$\displaystyle{=\frac{\hspace{2} {{\partial}}^{2} \hspace{2}}{\hspace{2} {\partial}x{\partial}y \hspace{2}}}$

$3$\displaystyle{ =\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}x \hspace{2}} \{ - { $ x- y $ }^{ - 1 } \} }$

$3$\displaystyle{ =- $ - 1 $ { $ x- y $ }^{ - 2 }\cdot 1 }$

$3$\displaystyle{={ $ x- y $ }^{ - 2 }}$

$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2} { $ x- y $ }^{2} \hspace{2}}}$ 　　･･････(2)

(1)，(2)より

$3$\displaystyle{\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}y{\partial}x \hspace{2}}=\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}x{\partial}y \hspace{2}}}$

となり，2次偏導関数は偏微分する順序には無関係であることが確かめられた．

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最終更新日： 2023年8月29日